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The sum of n terms of an AP that has a as the first term and the common difference is d is given by `(n/2)(2a + (n-1)*d)`
Here, the sum of the first 8 terms is 240
=> 240 = 4*(2a + 7d) ...(1)
And the sum of the first 16 terms is 1000
=> 1000 = 8(2a + 15d)
=> 500 = 4(2a + 15d) ...(2)
(2) - (1)
=> 260 = 60d - 28d
=> 260 = 32d
=> d = 8.125
240 = 4*(2a + 56.875)
=> 2a = 60 - 56.875
=> a = 1.5625
The 10th term is 1.5625 + 9*8.125 = 74.6875
The 10th term of the AP is 74.6875
In an AP series, the middle term is the average of the series.
So, sum is (average * no. of terms)
In case there are 8 terms, average will lie between 4th term and 5th term.
In case there are 16 terms, average will lie between 8th term and 9th term.
So, in the first case, 240 = (Average * 8)
So, average is 30.
In second case, 1000 = (Average * 16)
So, average is 62.5
In first case, average is 30 and in second case is average is 62.5, between these two averages, there is a gap of 4d ( d is the common difference) which is equal to 32.5.
We need to find 10th term of the series which is at a gap of 1.5d from the second average.
So, 4d = 32.5
Then 12d = 97.5
and then 1.5 d = 12.2
Thus the 10th term will be 62.5 + 12.2 = 74.7
(Average of second series is exactly between 8th and 9th term) and difference is of 1.5d between this average and 10th term.
so find out the magnitude of 1.5d and add it to the second average. we will get the 10th term of the series.
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